Calculus 8th Edition

$6\sqrt{3}$
$y=\sec^2 x$, $y=8\cos x$ Find the intersections: $\sec^2 x=8\cos x$ $\cos^2 x\sec^2 x=\cos^2 x*(8\cos x)$ $1=8\cos^3 x$ $\cos^3 x=\frac{1}{8}$ $\cos x=\frac{1}{2}$ $\cos x=\frac{\pi}{3}+2k\pi, -\frac{\pi}{3}+2k\pi$ It is given in the problem that we only want the area between $x=-\frac{\pi}{3}$ and $x=\frac{\pi}{3}$. Note that between $-\frac{\pi}{3}$ and $\frac{\pi}{3}$ $8\cos x>\sec^2 x$, so the integrand is $8\cos x-\sec^2 x$. The area is: $\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(8\cos x-\sec^2 x)dx$ $=(8\sin x-\tan x)\left|_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\right.$ $=(8\sin\frac{\pi}{3}-\tan\frac{\pi}{3})-(8\sin(-\frac{\pi}{3})-\tan(-\frac{\pi}{3}))$ $=(8*\frac{\sqrt{3}}{2}-\sqrt{3})-(-8*\frac{\sqrt{3}}{2}-(-\sqrt{3}))$ $=4\sqrt{3}-\sqrt{3}+4\sqrt{3}-\sqrt{3}$ $=6\sqrt{3}$